BdMO National 2012: Secondary 8

[Zzzz]

Problem 8:
The vertices of a right triangle $ABC$ inscribed in a circle divide the circumference into three arcs. The right angle is at $A$, so that the opposite arc $BC$ is a semicircle while arc $AB$ and arc $AC$ are supplementary. To each of the three arcs, we draw a tangent such that its point of tangency is the midpoint of that portion of the tangent intercepted by the extended lines $AB$ and $AC$. More precisely, the point $D$ on arc $BC$ is the midpoint of the segment joining the points $D'$ and $D''$ where the tangent at $D$ intersects the extended lines $AB$ and $AC$. Similarly for $E$ on arc $AC$ and $F$ on arc $AB$. Prove that triangle $DEF$ is equilateral.

[Tahmid Hasan]

let $D'=D_c$ and $D''=D_b$[a little change in the denotions]
let the tangent line at $F$ intersect $AB,AC$ at $F_c,F_b$ respectively.
let the tangent line at $E$ intersect $AB,AC$ at $E_c,E_b$ respectively.
now $DD_c=DD_b$ and $OD \bot D_cD_b$.so $OD_c=OD_b$ and $\angle DOD_c = \angle DOD_b$....(1)
again $DD_c=DD_b=D_cF=D_bE$
so $\triangle DFD_c \cong \triangle DED_b$[SSS]
hence $\angle FOD_c = \angle EOD_b$....(2)
adding (1),(2) we get $\angle FOD = \angle EOD$
which implies $\triangle FOD \cong \triangle EOD$[SAS]
so $FD=ED$
$\angle F_bAF_c = \angle E_cAE_B = 90^{\circ}$
so the circles wih diameter $F_bF_c,E_bE_c$ goes through $A$
which implies $FA=FF_b=FF_c,EA=EE_b=EE_c$
so $\angle FAB = \frac {1}{2} \angle AFF_b$
or,$\angle FDB = \frac {1}{2} \angle ADF$
similarly $\angle CDE = \frac {1}{2} \angle ADE$
now $\angle BDC = \angle CDE + \angle ADE + \angle ADF + \angle BDF$
$=\frac {1}{2} \angle ADE + \angle ADE + \angle ADF + \frac {1}{2} \angle ADF$
$= \frac {3}{2} \angle FDE$
so $\angle FDE = 60^{\circ}$
which implies $\triangle DEF$ is equilateral.
[1:10 hours,hmm... my timing has improved ]

[*Mahi*]

again $DD_c=DD_b=D_cF=D_bE$

I think I don't get the notations, because is this possible? What you said implies , from point $D_b$, $DD_b$ and $ED_b$ are the two tangents. Is that true? Please clarify.

[Tahmid Hasan]

I was revisiting old BdMO problems and I've a solution now!
$\angle A=90^{\circ} \Rightarrow DD'=DD"=DA,EE'=EE"=EA$.
$DA=DD" \Rightarrow \angle DAC=\angle DD"A$.
So $\angle ACD=\angle ADD'=2\angle DAC$.....$(1)$
Similarly we get $\angle ECA=2\angle EAC$.....$(2)$
$\angle DAE+\angle ECD=180^{\circ} \Rightarrow \angle DAE+2(\angle DAC+\angle ACD)$
$\Rightarrow 3\angle DAE=180^{\circ} \Rightarrow \angle DAE=\angle DFE=60^{\circ}$.
Similarly $\angle DEF=60^{\circ}$, So $\triangle DEF$ is equilateral.

[zadid xcalibured]

I don't know how the hell you deduced that $\angle ECA=2\angle EAC$.Please explain.
Your solution is incorrect.

[Tahmid Hasan]

I don't know how the hell you deduced that $\angle ECA=2\angle EAC$.Please explain.
Your solution is incorrect.

$EA=EE" \Rightarrow \angle EE"A=\angle EAC$
$\angle EE"A+\angle EAC=\angle AEE' \Rightarrow 2\angle EAC=\angle ACE$.
I hope it's clear now.

[zadid xcalibured]

I don't know how the hell you deduced $EA=EE"$
Because that implies $\angle E'AE"$ $=90 ^\circ$ which is not true.

[Tahmid Hasan]

I don't know how the hell you deduced $EA=EE"$
Because that implies $\angle E'AE"$ $=90 ^\circ$ which is not true.

Actually it is . Denote $\ell_X$ by the tangent drawn at $X$.
According to the problem statement $E'=AB \cap \ell_E,E"=AC \cap \ell_E,F'=AB \cap \ell_F,F"=AC \cap \ell_F$
I also misinterpreted it in the actual exam thinking $E',E"$ as the intersection of $\ell_E$ with $AB,BC$ and likewise. Here's a figure for proper understanding, cheers

[zadid xcalibured]

That's what stops the good ninja(me).Mis-comprehension of problem statement.
Now i know how the hell u deduce those things.

[Tahmid Hasan]

Read the question again,brother.The definition for all $E$ and $F$ are similar as $D$.At least the question says so.

I don't think so.

To each of the three arcs, we draw a tangent such that its point of tangency is the midpoint of that portion of the tangent intercepted by the extended lines $AB$ and $AC$.

See, the intersections have nothing to do with $BC$; only $AB,AC$ for 'all' three points.